Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x-5y &= -5 \\ -8x+6y &= 2\end{align*}$
Answer: Begin by moving the $x$ -term in the second equation to the right side of the equation. $6y = 8x+2$ Divide both sides by $6$ to isolate $y$ $y = {\dfrac{4}{3}x + \dfrac{1}{3}}$ Substitute this expression for $y$ in the first equation. $3x-5({\dfrac{4}{3}x + \dfrac{1}{3}}) = -5$ $3x - \dfrac{20}{3}x - \dfrac{5}{3} = -5$ Simplify by combining terms, then solve for $x$ $-\dfrac{11}{3}x - \dfrac{5}{3} = -5$ $-\dfrac{11}{3}x = -\dfrac{10}{3}$ $x = \dfrac{10}{11}$ Substitute $\dfrac{10}{11}$ for $x$ back into the top equation. $3( \dfrac{10}{11})-5y = -5$ $\dfrac{30}{11}-5y = -5$ $-5y = -\dfrac{85}{11}$ $y = \dfrac{17}{11}$ The solution is $\enspace x = \dfrac{10}{11}, \enspace y = \dfrac{17}{11}$.